just as light does. Because the wavelength of electrons can be much smaller than that ofvisible light, much greater resolution and magnification can be obtained. A scanningelectron microscope (SEM) can produce images with a three-dimensional quality. All EM images are monochromatic (black and white). Artistic coloring has beenadded here, as is common. On the left is an SEM image of a blood clot forming(yellow-color web) due to a wound. White blood cells are colored green here forvisibility. On the right, red blood cells in a small artery. A red blood cell travels about15 km a day inside our bodies and lives roughly 4 months before damage or rupture.Humans contain 4 to 6 liters of blood, and 2 to 3 * 1013 red blood cells.

A P T E H

27 R CEarly Quantum Theoryand Models of the AtomCHAPTER-OPENING QUESTIONGuess now! CONTENTSIt has been found experimentally that 271 Discovery and Properties (a) light behaves as a wave. of the Electron (b) light behaves as a particle. 272 Blackbody Radiation; (c) electrons behave as particles. Plancks Quantum Hypothesis (d) electrons behave as waves. 273 Photon Theory of Light and the Photoelectric Effect (e) all of the above are true. 274 Energy, Mass, and (f) only (a) and (b) are true. Momentum of a Photon (g) only (a) and (c) are true. *275 Compton Effect (h) none of the above are true. 276 Photon Interactions; Pair Production

T he second aspect of the revolution that shook the world of physics in the 277 WaveParticle Duality; the early part of the twentieth century was the quantum theory (the other was Principle of Complementarity Einsteins theory of relativity). Unlike the special theory of relativity, the 278 Wave Nature of Matterrevolution of quantum theory required almost three decades to unfold, and many 279 Electron Microscopesscientists contributed to its development. It began in 1900 with Plancks quantum 2710 Early Models of the Atomhypothesis, and culminated in the mid-1920s with the theory of quantum 2711 Atomic Spectra: Key to themechanics of Schrdinger and Heisenberg which has been so effective in explain- Structure of the Atoming the structure of matter. The discovery of the electron in the 1890s, with which 2712 The Bohr Modelwe begin this Chapter, might be said to mark the beginning of modern physics, 2713 de Broglies Hypothesisand is a sort of precursor to the quantum theory. Applied to Atoms

771 271 Discovery and Properties of the Electron Screens Toward the end of the nineteenth century, studies were being done on the Glow discharge of electricity through rarefied gases. One apparatus, diagrammed in Cathode Fig. 271, was a glass tube fitted with electrodes and evacuated so only a small amount of gas remained inside. When a very high voltage was applied to the Anode electrodes, a dark space seemed to extend outward from the cathode (negative High + electrode) toward the opposite end of the tube; and that far end of the tube would voltage glow. If one or more screens containing a small hole were inserted as shown,FIGURE 271 Discharge tube. In the glow was restricted to a tiny spot on the end of the tube. It seemed as thoughsome models, one of the screens is something being emitted by the cathode traveled across to the opposite end of thethe anode (positive plate). tube. These somethings were named cathode rays. There was much discussion at the time about what these rays might be. Some scientists thought they might resemble light. But the observation that the bright spot at the end of the tube could be deflected to one side by an electric or magnetic field suggested that cathode rays were charged particles; and the direction of the deflection was consistent with a negative charge. Furthermore, if the tube con- tained certain types of rarefied gas, the path of the cathode rays was made visible by a slight glow. Estimates of the charge e of the cathode-ray particles, as well as of their charge-to-mass ratio em, had been made by 1897. But in that year, J. J. Thomson (18561940) was able to measure em directly, using the apparatus shown in Fig. 272. Cathode rays are accelerated by a high voltage and then pass between a pair of parallel plates built into the tube. Another voltage applied to the B parallel plates produces an electric field E, and a pair of coils produces a B magnetic field B. If E = B = 0, the cathode rays follow path b in Fig. 272. + I a I Anode + +FIGURE 272 Cathode rays +deflected by electric and magnetic + bfields. (See also Section 1711 on the CRT.)

c High voltage Electric field plates Coils to produce magnetic field

When only the electric field is present, say with the upper plate positive, the cathode rays are deflected upward as in path a in Fig. 272. If only a magnetic field exists, say inward, the rays are deflected downward along path c. These observations are just what is expected for a negatively charged particle. The force on the rays due to the magnetic field is F = evB, where e is the charge and v is the velocity of the cathode rays (Eq. 204). In the absence of an electric field, the rays are bent into a curved path, and applying Newtons second law F = ma with a = centripetal acceleration gives v2 , evB = m r and thus e v . = m Br The radius of curvature r can be measured and so can B. The velocity v can be found by applying an electric field in addition to the magnetic field. The electric

772 CHAPTER 27 Early Quantum Theory and Models of the Atom

field E is adjusted so that the cathode rays are undeflected and follow path b inFig. 272. In this situation the upward force due to the electric field, F = eE, isbalanced by the downward force due to the magnetic field, F = evB. We equatethe two forces, eE = evB, and find E. v = BCombining this with the above equation we have e E . = (27;1) m B2rThe quantities on the right side can all be measured, and although e and m couldnot be determined separately, the ratio em could be determined. The acceptedvalue today is em = 1.76 * 1011 Ckg. Cathode rays soon came to be calledelectrons.

Discovery in ScienceThe discovery of the electron, like many others in science, is not quite soobvious as discovering gold or oil. Should the discovery of the electron be creditedto the person who first saw a glow in the tube? Or to the person who first calledthem cathode rays? Perhaps neither one, for they had no conception of the electronas we know it today. In fact, the credit for the discovery is generally given toThomson, but not because he was the first to see the glow in the tube. Rather itis because he believed that this phenomenon was due to tiny negatively chargedparticles and made careful measurements on them. Furthermore he argued thatthese particles were constituents of atoms, and not ions or atoms themselves asmany thought, and he developed an electron theory of matter. His view is close towhat we accept today, and this is why Thomson is credited with the discovery.Note, however, that neither he nor anyone else ever actually saw an electron itself.We discuss this briefly, for it illustrates the fact that discovery in science is notalways a clear-cut matter. In fact some philosophers of science think the worddiscovery is often not appropriate, such as in this case.

Electron Charge Measurement Atomizer

Thomson believed that an electron was not an atom, but rather a constituent,or part, of an atom. Convincing evidence for this came soon with the determin-ation of the charge and the mass of the cathode rays. Thomsons student +J. S. Townsend made the first direct (but rough) measurements of e in 1897. But + + +it was the more refined oil-drop experiment of Robert A. Millikan (18681953) Dropletsthat yielded a precise value for the charge on the electron and showed that charge Telescopecomes in discrete amounts. In this experiment, tiny droplets of mineral oil carrying

an electric charge were allowed to fall under gravity between two parallel plates,Fig. 273. The electric field E between the plates was adjusted until the drop was FIGURE 273 Millikans oil-dropsuspended in midair. The downward pull of gravity, mg, was then just balanced by experiment.the upward force due to the electric field. Thus qE = mg so the charge q = mgE.The mass of the droplet was determined by measuring its terminal velocity in theabsence of the electric field. Often the droplet was charged negatively, but some-times it was positive, suggesting that the droplet had acquired or lost electrons (byfriction, leaving the atomizer). Millikans painstaking observations and analysis pre-sented convincing evidence that any charge was an integral multiple of a smallestcharge, e, that was ascribed to the electron, and that the value of e was 1.6 * 1019 C.This value of e, combined with the measurement of em, gives the mass of theelectron to be A1.6 * 10 19 CBA1.76 * 1011 CkgB = 9.1 * 1031 kg. This mass isless than a thousandth the mass of the smallest atom, and thus confirmed the ideathat the electron is only a part of an atom. The accepted value today for the massof the electron is me = 9.11 * 10 31 kg.The experimental result that any charge is an integral multiple of e means thatelectric charge is quantized (exists only in discrete amounts). SECTION 271 773 272 Blackbody Radiation; Plancks Quantum Hypothesis Blackbody Radiation One of the observations that was unexplained at the end of the nineteenth cen- tury was the spectrum of light emitted by hot objects. We saw in Section 148 that all objects emit radiation whose total intensity is proportional to the fourth power of the Kelvin (absolute) temperature AT4 B. At normal temperatures (L 300 K), we are not aware of this electromagnetic radiation because of its low Frequency (Hz) intensity. At higher temperatures, there is sufficient infrared radiation that we 1.0 3.0 2.0 1.0 can feel heat if we are close to the object. At still higher temperatures (on the 1015 1014 1014 1014 order of 1000 K), objects actually glow, such as a red-hot electric stove burner or the heating element in a toaster. At temperatures above 2000 K, objects glow with a yellow or whitish color, such as white-hot iron and the filament of a lightbulb. The light emitted contains a continuous range of wavelengths or frequencies, and the spectrum is a plot of intensity vs. wavelength or frequency. As the temperatureIntensity

6000 K increases, the electromagnetic radiation emitted by objects not only increases in total intensity but has its peak intensity at higher and higher frequencies. 4500 K The spectrum of light emitted by a hot dense object is shown in Fig. 274 for an idealized blackbody. A blackbody is a body that, when cool, would absorb all 3000 K the radiation falling on it (and so would appear black under reflection when illuminated by other sources). The radiation such an idealized blackbody would 0 UV 1000 IR 2000 3000 emit when hot and luminous, called blackbody radiation (though not necessarily Visible Wavelength (nm) black in color), approximates that from many real objects. The 6000-K curve in FIGURE 274 Measured spectra of Fig. 274, corresponding to the temperature of the surface of the Sun, peaks in wavelengths and frequencies the visible part of the spectrum. For lower temperatures, the total intensity drops emitted by a blackbody at three considerably and the peak occurs at longer wavelengths (or lower frequencies). different temperatures. This is why objects glow with a red color at around 1000 K. It is found experimen- tally that the wavelength at the peak of the spectrum, lP , is related to the Kelvin temperature T by lP T = 2.90 * 10 3 mK. (27;2) This is known as Wiens law. EXAMPLE 27;1 The Suns surface temperature. Estimate the tempera- ture of the surface of our Sun, given that the Sun emits light whose peak intensity occurs in the visible spectrum at around 500 nm. APPROACH We assume the Sun acts as a blackbody, and use lP = 500 nm in Wiens law (Eq. 272). SOLUTION Wiens law gives 2.90 * 10 3 mK 2.90 * 103 mK T = = L 6000 K. lP 500 * 109 m

EXAMPLE 27;2 Star color. Suppose a star has a surface temperature of

32,500 K. What color would this star appear? APPROACH We assume the star emits radiation as a blackbody, and solve for lP in Wiens law, Eq. 272. SOLUTION From Wiens law we have 2.90 * 10 3 mK 2.90 * 10 3 mK lP = = = 89.2 nm. T 3.25 * 104 K The peak is in the UV range of the spectrum, and will be way to the left in Fig. 274. In the visible region, the curve will be descending, so the shortest visible wavelengths will be strongest. Hence the star will appear bluish (or blue-white). NOTE This example helps us to understand why stars have different colors (reddish for the coolest stars; orangish, yellow, white, bluish for hotter stars.)

EXERCISE A What is the color of an object at 4000 K?

774 CHAPTER 27Plancks Quantum HypothesisIn the year 1900, Max Planck (18581947) proposed a theory that was able toreproduce the graphs of Fig. 274. His theory, still accepted today, made a new andradical assumption: that the energy of the oscillations of atoms within moleculescannot have just any value; instead each has energy which is a multiple of a mini-mum value related to the frequency of oscillation by E = hf.Here h is a new constant, now called Plancks constant, whose value wasestimated by Planck by fitting his formula for the blackbody radiation curve toexperiment. The value accepted today is h = 6.626 * 1034 Js.Plancks assumption suggests that the energy of any molecular vibration could beonly a whole number multiple of hf : E = nhf, n = 1, 2, 3, p , (27;3)where n is called a quantum number (quantum means discrete amount asopposed to continuous). This idea is often called Plancks quantum hypothesis,although little attention was brought to this point at the time. In fact, it appearsthat Planck considered it more as a mathematical device to get the right answerrather than as an important discovery. Planck himself continued to seek a classicalexplanation for the introduction of h. The recognition that this was an important (a)and radical innovation did not come until later, after about 1905 when others,particularly Einstein, entered the field. The quantum hypothesis, Eq. 273, states that the energy of an oscillatorcan be E = hf, or 2hf, or 3hf, and so on, but there cannot be vibrations withenergies between these values. That is, energy would not be a continuous quan-tity as had been believed for centuries; rather it is quantizedit exists onlyin discrete amounts. The smallest amount of energy possible (hf) is called thequantum of energy. Recall from Chapter 11 that the energy of an oscillation isproportional to the amplitude squared. Another way of expressing the quantum (b)hypothesis is that not just any amplitude of vibration is possible. The possible FIGURE 275 Ramp versus stairvalues for the amplitude are related to the frequency f . analogy. (a) On a ramp, a box can have A simple analogy may help. Compare a ramp, on which a box can be placed continuous values of potential energy.at any height, to a flight of stairs on which the box can have only certain discrete (b) But on stairs, the box can haveamounts of potential energy, as shown in Fig. 275. only discrete (quantized) values of energy.

273 Photon Theory of Light and

the Photoelectric EffectIn 1905, the same year that he introduced the special theory of relativity, Einsteinmade a bold extension of the quantum idea by proposing a new theory of light.Plancks work had suggested that the vibrational energy of molecules in a radiat-ing object is quantized with energy E = nhf, where n is an integer and f is thefrequency of molecular vibration. Einstein argued that when light is emitted by amolecular oscillator, the molecules vibrational energy of nhf must decrease by anamount hf (or by 2hf, etc.) to another integer times hf, such as (n - 1)hf. Thento conserve energy, the light ought to be emitted in packets, or quanta, each withan energy E = hf, (27;4) Photon energy

where f is here the frequency of the emitted light. Again h is Plancks constant.Because all light ultimately comes from a radiating source, this idea suggests thatlight is transmitted as tiny particles, or photons as they are now called, as well as viathe waves predicted by Maxwells electromagnetic theory. The photon theory oflight was also a radical departure from classical ideas. Einstein proposed a test of thequantum theory of light: quantitative measurements on the photoelectric effect.

SECTION 273 Photon Theory of Light and the Photoelectric Effect 775 Light When light shines on a metal surface, electrons are found to be emitted from source the surface. This effect is called the photoelectric effect and it occurs in many materials, but is most easily observed with metals. It can be observed using the apparatus shown in Fig. 276. A metal plate P and a smaller electrode C are placed inside an evacuated glass tube, called a photocell. The two electrodes are Light connected to an ammeter and a source of emf, as shown. When the photocell is in the dark, the ammeter reads zero. But when light of sufficiently high frequency illuminates the plate, the ammeter indicates a current flowing in the circuit. We C P explain completion of the circuit by imagining that electrons, ejected from the plate by the impinging light, flow across the tube from the plate to the collector C as indicated in Fig. 276. Photocell That electrons should be emitted when light shines on a metal is consistent A with the electromagnetic (EM) wave theory of light: the electric field of an EM wave could exert a force on electrons in the metal and eject some of them. V Einstein pointed out, however, that the wave theory and the photon theory of + light give very different predictions on the details of the photoelectric effect. ForFIGURE 276 The photoelectric example, one thing that can be measured with the apparatus of Fig. 276 is theeffect. maximum kinetic energy A kemax B of the emitted electrons. This can be done by using a variable voltage source and reversing the terminals so that electrode C is negative and P is positive. The electrons emitted from P will be repelled by the negative electrode, but if this reverse voltage is small enough, the fastest electrons will still reach C and there will be a current in the circuit. If the reversed voltage is increased, a point is reached where the current reaches zerono electrons have sufficient kinetic energy to reach C. This is called the stopping potential, or stopping voltage, V0 , and from its measurement, kemax can be determined using conservation of energy (loss of kinetic energy = gain in potential energy): kemax = eV0 . Now let us examine the details of the photoelectric effect from the point of view of the wave theory versus Einsteins particle theory. First the wave theory, assuming monochromatic light. The two important properties of a light wave are its intensity and its frequency (or wavelength). When these two quantities are varied, the wave theory makes the following predictions: Wave 1. If the light intensity is increased, the number of electrons ejected and their maximum kinetic energy should be increased because the higher intensity theory means a greater electric field amplitude, and the greater electric field should eject electrons with higher speed. predictions 2. The frequency of the light should not affect the kinetic energy of the ejected electrons. Only the intensity should affect kemax . The photon theory makes completely different predictions. First we note that in a monochromatic beam, all photons have the same energy (= hf). Increasing the intensity of the light beam means increasing the number of photons in the beam, but does not affect the energy of each photon as long as the frequency is not changed. According to Einsteins theory, an electron is ejected from the metal by a collision with a single photon. In the process, all the photon energy is trans- ferred to the electron and the photon ceases to exist. Since electrons are held in the metal by attractive forces, some minimum energy W0 is required just to get an electron out through the surface. W0 is called the work function, and is a few electron volts A1 eV = 1.6 * 1019 JB for most metals. If the frequency f of the incoming light is so low that hf is less than W0 , then the photons will not have enough energy to eject any electrons at all. If hf 7 W0 , then electrons will be ejected and energy will be conserved in the process. That is, the input energy (of the photon), hf, will equal the outgoing kinetic energy ke of the electron plus the energy required to get it out of the metal, W: hf = ke + W. (27;5a) The least tightly held electrons will be emitted with the most kinetic energy A kemax B,

776 CHAPTER 27 Early Quantum Theory and Models of the Atom

in which case W in this equation becomes the work function W0 , and kebecomes kemax : hf = kemax + W0 . [least bound electrons] (27;5b)Many electrons will require more energy than the bare minimum AW0 B to get outof the metal, and thus the kinetic energy of such electrons will be less than themaximum. From these considerations, the photon theory makes the following predictions: 1. An increase in intensity of the light beam means more photons are incident, so more electrons will be ejected; but since the energy of each photon is not Photon changed, the maximum kinetic energy of electrons is not changed by an increase in intensity. theory 2. If the frequency of the light is increased, the maximum kinetic energy of the electrons increases linearly, according to Eq. 275b. That is, kemax = hf - W0 . predictions

This relationship is plotted in Fig. 277.

3. If the frequency f is less than the cutoff frequency f0 , where hf0 = W0 , no

of electrons electrons will be ejected, no matter how great the intensity of the light.

KE max These predictions of the photon theory are very different from the predictionsof the wave theory. In 19131914, careful experiments were carried out by R. A.Millikan. The results were fully in agreement with Einsteins photon theory. One other aspect of the photoelectric effect also confirmed the photon f0 Frequency of light ftheory. If extremely low light intensity is used, the wave theory predicts a time FIGURE 277 Photoelectric effect:delay before electron emission so that an electron can absorb enough energy to the maximum kinetic energy ofexceed the work function. The photon theory predicts no such delayit only ejected electrons increases linearlytakes one photon (if its frequency is high enough) to eject an electronand with the frequency of incident light.experiments showed no delay. This too confirmed Einsteins photon theory. No electrons are emitted if f 6 f0 .

EXAMPLE 27;4 ESTIMATE Photons from a lightbulb. Estimate how

many visible light photons a 100-W lightbulb emits per second. Assume the bulb has a typical efficiency of about 3% (that is, 97% of the energy goes to heat). APPROACH Lets assume an average wavelength in the middle of the visible spectrum, l L 500 nm. The energy of each photon is E = hf = hcl. Only 3% of the 100-W power is emitted as visible light, or 3 W = 3 Js. The number of photons emitted per second equals the light output of 3 Js divided by the energy of each photon. SOLUTION The energy emitted in one second (= 3 J) is E = Nhf where N is the number of photons emitted per second and f = cl. Hence E El (3 J)A500 * 10 9 mB N = = = L 8 * 1018 hf hc A6.63 * 10 34 JsBA3.00 * 108 msB per second, or almost 1019 photons emitted per second, an enormous number.

SECTION 273 Photon Theory of Light and the Photoelectric Effect 777 EXERCISE B A beam contains infrared light of a single wavelength, 1000 nm, and monochromatic UV at 100 nm, both of the same intensity. Are there more 100-nm photons or more 1000-nm photons?

EXAMPLE 27;5 Photoelectron speed and energy. What is the kinetic

energy and the speed of an electron ejected from a sodium surface whose work function is W0 = 2.28 eV when illuminated by light of wavelength (a) 410 nm, (b) 550 nm? APPROACH We first find the energy of the photons (E = hf = hcl). If the energy is greater than W0 , then electrons will be ejected with varying amounts of ke, with a maximum of kemax = hf - W0 . SOLUTION (a) For l = 410 nm, hc hf = = 4.85 * 10 19 J or 3.03 eV. l The maximum kinetic energy an electron can have is given by Eq. 275b, kemax = 3.03 eV - 2.28 eV = 0.75 eV, or (0.75 eV)(1.60 * 1019 JeV) = 1.2 * 10 19 J. Since ke = 12 mv2 where m = 9.1 * 10 31 kg, 2ke vmax = = 5.1 * 105 ms. B m Most ejected electrons will have less ke and less speed than these maximum values. (b) For l = 550 nm, hf = hcl = 3.61 * 10 19 J = 2.26 eV. Since this photon energy is less than the work function, no electrons are ejected. NOTE In (a) we used the nonrelativistic equation for kinetic energy. If v had turned out to be more than about 0.1c, our calculation would have been inaccurate by more than a percent or so, and we would probably prefer to redo it using the relativistic form (Eq. 265).

EXERCISE C Determine the lowest frequency and the longest wavelength needed to emit electrons from sodium. By converting units, we can show that the energy of a photon in electron volts, when given the wavelength l in nm, is 1.240 * 103 eVnm . E (eV) = [photon energy in eV] l (nm)

Applications of the Photoelectric Effect

The photoelectric effect, besides playing an important historical role in confirm- ing the photon theory of light, also has many practical applications. Burglar alarms FIGURE 278 Optical sound track and automatic doors often make use of the photocell circuit of Fig. 276. When on movie film. In the projector, light a person interrupts the beam of light, the sudden drop in current in the circuit from a small source (different from that for the picture) passes through activates a switchoften a solenoidwhich operates a bell or opens the door. the sound track on the moving film. UV or IR light is sometimes used in burglar alarms because of its invisibility. Many smoke detectors use the photoelectric effect to detect tiny amounts of smoke Picture Sound track that interrupt the flow of light and so alter the electric current. Photographic light meters use this circuit as well. Photocells are used in many other devices, such as Photocell absorption spectrophotometers, to measure light intensity. One type of film sound track is a variably shaded narrow section at the side of the film, Fig. 278. Light passing through the film is thus modulated, and the output electrical signal of the photocell detector follows the frequencies on the sound track. For manySmall applications today, the vacuum-tube photocell of Fig. 276 has been replaced by alightsource semiconductor device known as a photodiode (Section 299). In these semicon- ductors, the absorption of a photon liberates a bound electron so it can move freely, which changes the conductivity of the material and the current through a photodiode is altered.

778 CHAPTER 27 Early Quantum Theory and Models of the Atom

274 Energy, Mass, and Momentum of a PhotonWe have just seen (Eq. 274) that the total energy of a single photon is given byE = hf. Because a photon always travels at the speed of light, it is truly a rela-tivistic particle. Thus we must use relativistic formulas for dealing with its mass,energy, and momentum. The momentum of any particle of mass m is given byp = mv31 - v2c2. Since v = c for a photon, the denominator is zero. Toavoid having an infinite momentum, we conclude that the photons mass must bezero: m = 0. This makes sense too because a photon can never be at rest (italways moves at the speed of light). A photons kinetic energy is its total energy: ke = E = hf. [photon]The momentum of a photon can be obtained from the relativistic formula(Eq. 269) E 2 = p2c2 + m2c4 where we set m = 0, so E 2 = p2c2 or E. p = [photon] CAUTION c Momentum of photon is not mvSince E = hf for a photon, its momentum is related to its wavelength by E hf h. p = = = (27;6) c c l

EXAMPLE 27;6 ESTIMATE Photon momentum and force. Suppose the

1019 photons emitted per second from the 100-W lightbulb in Example 274 were all focused onto a piece of black paper and absorbed. (a) Calculate the momentum of one photon and (b) estimate the force all these photons could exert on the paper. APPROACH Each photons momentum is obtained from Eq. 276, p = hl. Next, each absorbed photons momentum changes from p = hl to zero. We use Newtons second law, F = p t, to get the force. Let l = 500 nm. SOLUTION (a) Each photon has a momentum h 6.63 * 10 34 Js p = = = 1.3 * 10 27 kgms. l 500 * 10 9 m (b) Using Newtons second law for N = 1019 photons (Example 274) whose momentum changes from hl to 0, we obtain p Nhl - 0 h F = = = N L A1019 s1 BA1027 kgmsB L 10 8 N. t 1s l NOTE This is a tiny force, but we can see that a very strong light source could exert a measurable force, and near the Sun or a star the force due to photons in electromagnetic radiation could be considerable. See Section 226.

EXAMPLE 27;7 Photosynthesis. In photosynthesis, pigments such as PHYSICS APPLIED

chlorophyll in plants capture the energy of sunlight to change CO2 to useful Photosynthesis carbohydrate. About nine photons are needed to transform one molecule of CO2 to carbohydrate and O2 . Assuming light of wavelength l = 670 nm (chlorophyll absorbs most strongly in the range 650 nm to 700 nm), how efficient is the photosynthetic process? The reverse chemical reaction releases an energy of 4.9 eVmolecule of CO2 , so 4.9 eV is needed to transform CO2 to carbohydrate. APPROACH The efficiency is the minimum energy required (4.9 eV) divided by the actual energy absorbed, nine times the energy (hf) of one photon. SOLUTION The energy of nine photons, each of energy hf = hcl, is (9)A6.63 * 1034 JsBA3.00 * 108 msBA6.7 * 10 7 mB = 2.7 * 1018 J or 17 eV. Thus the process is about (4.9 eV17 eV) = 29% efficient.

SECTION 274 Energy, Mass, and Momentum of a Photon 779

* 275 Compton Effect BEFORE AFTER Besides the photoelectric effect, a number of other experiments were carried out COLLISION COLLISION in the early twentieth century which also supported the photon theory. One of y Scattered these was the Compton effect (1923) named after its discoverer, A. H. Compton photon (') (18921962). Compton aimed short-wavelength light (actually X-rays) at various Incident materials, and detected light scattered at various angles. He found that the scattered photon () light had a slightly longer wavelength than did the incident light, and therefore a x slightly lower frequency indicating a loss of energy. He explained this result on the Electron at rest basis of the photon theory as incident photons colliding with electrons of the initially e material, Fig. 279. Using Eq. 276 for momentum of a photon, Compton applied the laws of conservation of momentum and energy to the collision of Fig. 279FIGURE 279 The Compton effect. and derived the following equation for the wavelength of the scattered photons:A single photon of wavelength lstrikes an electron in some material, h l = l + (1 - cos f), (27;7)knocking it out of its atom. The mecscattered photon has less energy where me is the mass of the electron. (The quantity hmec, which has the dimen-(some energy is given to the sions of length, is called the Compton wavelength of the electron.) We see thatelectron) and hence has a longerwavelength l (shown exaggerated). the predicted wavelength of scattered photons depends on the angle f at whichExperiments found scattered X-rays they are detected. Comptons measurements of 1923 were consistent with thisof just the wavelengths predicted by formula. The wave theory of light predicts no such shift: an incoming electro-conservation of energy and magnetic wave of frequency f should set electrons into oscillation at frequency f;momentum using the photon model. and such oscillating electrons would reemit EM waves of this same frequency f (Section 222), which would not change with angle (f). Hence the Compton effect adds to the firm experimental foundation for the photon theory of light. EXERCISE D When a photon scatters off an electron by the Compton effect, which of the following increases: its energy, frequency, wavelength?

PHYSICS APPLIED The Compton effect has been used to diagnose bone disease such as osteoporo- Measuring bone density sis. Gamma rays, which are photons of even shorter wavelength than X-rays, coming from a radioactive source are scattered off bone material. The total intensity of the scattered radiation is proportional to the density of electrons, which is in turn proportional to the bone density. A low bone density may indicate osteoporosis.

780 CHAPTER 27 Early Quantum Theory and Models of the Atom

276 Photon Interactions; Pair ProductionWhen a photon passes through matter, it interacts with the atoms and electrons.There are four important types of interactions that a photon can undergo: 1. The photoelectric effect: A photon may knock an electron out of an atom and in the process the photon disappears. 2. The photon may knock an atomic electron to a higher energy state in the atom if its energy is not sufficient to knock the electron out altogether. In this process e the photon also disappears, and all its energy is given to the atom. Such an atom is then said to be in an excited state, and we shall discuss it more later. 3. The photon can be scattered from an electron (or a nucleus) and in the Photon process lose some energy; this is the Compton effect (Fig. 279). But notice + that the photon is not slowed down. It still travels with speed c, but its frequency will be lower because it has lost some energy. Nucleus

4. Pair production: A photon can actually create matter, such as the production e of an electron and a positron, Fig. 2710. (A positron has the same mass as FIGURE 2710 Pair production: an electron, but the opposite charge, e. ) a photon disappears and produces an electron and a positron.In process 4, pair production, the photon disappears in the process of creatingthe electronpositron pair. This is an example of mass being created from pureenergy, and it occurs in accord with Einsteins equation E = mc2. Notice that aphoton cannot create an electron alone since electric charge would not then beconserved. The inverse of pair production also occurs: if a positron comes close toan electron, the two quickly annihilate each other and their energy, includingtheir mass, appears as electromagnetic energy of photons. Because positrons arenot as plentiful in nature as electrons, they usually do not last long. Electronpositron annihilation is the basis for the type of medical imagingknown as PET, as discussed in Section 318. EXAMPLE 27;9 Pair production. (a) What is the minimum energy of a photon that can produce an electronpositron pair? (b) What is this photons wavelength? APPROACH The minimum photon energy E equals the rest energy Amc2 B of the two particles created, via Einsteins famous equation E = mc2 (Eq. 267). There is no energy left over, so the particles produced will have zero kinetic energy. The wavelength is l = cf where E = hf for the original photon. SOLUTION (a) Because E = mc2 , and the mass created is equal to two electron masses, the photon must have energy E = 2A9.11 * 10 31 kgBA3.00 * 108 msB 2 = 1.64 * 10 13 J = 1.02 MeV (1 MeV = 106 eV = 1.60 * 10 13 J). A photon with less energy cannot undergo pair production. (b) Since E = hf = hcl, the wavelength of a 1.02-MeV photon is hc A6.63 * 10 34 JsBA3.00 * 108 msB l = = = 1.2 * 10 12 m, E A1.64 * 10 13 JB which is 0.0012 nm. Such photons are in the gamma-ray (or very short X-ray) region of the electromagnetic spectrum (Fig. 228). NOTE Photons of higher energy (shorter wavelength) can also create an electron positron pair, with the excess energy becoming kinetic energy of the particles.

Pair production cannot occur in empty space, for momentum could not be con-served. In Example 279, for instance, energy is conserved, but only enough energywas provided to create the electronpositron pair at rest and thus with zero momen-tum, which could not equal the initial momentum of the photon. Indeed, it can beshown that at any energy, an additional massive object, such as an atomic nucleus(Fig. 2710), must take part in the interaction to carry off some of the momentum.

SECTION 276 Photon Interactions; Pair Production 781

277 WaveParticle Duality; the Principle of Complementarity The photoelectric effect, the Compton effect, and other experiments have placed the particle theory of light on a firm experimental basis. But what about the classic experiments of Young and others (Chapter 24) on interference and diffraction which showed that the wave theory of light also rests on a firm experimental basis? We seem to be in a dilemma. Some experiments indicate that light behaves like a wave; others indicate that it behaves like a stream of particles. These two theories seem to be incompatible, but both have been shown to have validity. Physicists finally came to the conclusion that this duality of light must be accepted as a fact of life. It is referred to as the wave;particle duality. Apparently, light is a more complex phenomenon than just a simple wave or a simple beam of particles. To clarify the situation, the great Danish physicist Niels Bohr (18851962, Fig. 2711) proposed his famous principle of complementarity. It states that to understand an experiment, sometimes we find an explanation using wave theory and sometimes using particle theory. Yet we must be aware of both the wave and FIGURE 2711 Niels Bohr (right), particle aspects of light if we are to have a full understanding of light. Therefore walking with Enrico Fermi along the Appian Way outside Rome. This these two aspects of light complement one another. photo shows one important way It is not easy to visualize this duality. We cannot readily picture a combina- physics is done. tion of wave and particle. Instead, we must recognize that the two aspects of light are different faces that light shows to experimenters. Part of the difficulty stems from how we think. Visual pictures (or models) in our minds are based on what we see in the everyday world. We apply the concepts of waves and particles to light because in the macroscopic world we see that energy is transferred from place to place by these two methods. We cannot see directly whether light is a wave or particle, so we do indirect experiments. To explain the experiments, we apply the models of waves or of particles to the nature of light. But these are abstractions of the human mind. When we try to conceive of what light really is, we insist on a visual picture. Yet there is no CAUTION reason why light should conform to these models (or visual images) taken fromNot correct to say light is a wave and/or the macroscopic world. The true nature of lightif that means anythingisa particle. Light can act like a wave or not possible to visualize. The best we can do is recognize that our knowledge is like a particle limited to the indirect experiments, and that in terms of everyday language and images, light reveals both wave and particle properties. It is worth noting that Einsteins equation E = hf itself links the particle and wave properties of a light beam. In this equation, E refers to the energy of a particle; and on the other side of the equation, we have the frequency f of the corresponding wave.

278 Wave Nature of Matter

In 1923, Louis de Broglie (18921987) extended the idea of the waveparticle duality. He appreciated the symmetry in nature, and argued that if light some- times behaves like a wave and sometimes like a particle, then perhaps those things in nature thought to be particlessuch as electrons and other material objects might also have wave properties. De Broglie proposed that the wavelength of a material particle would be related to its momentum in the same way as for a photon, Eq. 276, p = hl. That is, for a particle having linear momentum p = mv, the wavelength l is given by h, de Broglie wavelength l = (27;8) p

Ordinary objects, such as the ball of Example 2710, have unimaginably smallwavelengths. Even if the speed is extremely small, say 104 ms, the wavelengthwould be about 10 29 m. Indeed, the wavelength of any ordinary object is muchtoo small to be measured and detected. The problem is that the properties ofwaves, such as interference and diffraction, are significant only when the size ofobjects or slits is not much larger than the wavelength. And there are no knownobjects or slits to diffract waves only 10 30 m long, so the wave properties ofordinary objects go undetected. But tiny elementary particles, such as electrons, are another matter. Sincethe mass m appears in the denominator of Eq. 278, a very small mass shouldhave a much larger wavelength. EXAMPLE 27;11 Wavelength of an electron. Determine the wavelength of an electron that has been accelerated through a potential difference of 100 V. APPROACH If the kinetic energy is much less than the rest energy, we can use the classical formula, ke = 12 mv2 (see end of Section 269). For an electron, mc2 = 0.511 MeV. We then apply conservation of energy: the kinetic energy acquired by the electron equals its loss in potential energy. After solving for v, we use Eq. 278 to find the de Broglie wavelength. SOLUTION The gain in kinetic energy equals the loss in potential energy: pe = eV - 0. Thus ke = eV, so ke = 100 eV. The ratio kemc2 = 100 eVA0.511 * 106 eVB L 104 , so relativity is not needed. Thus 1 mv2 = eV 2 and 2 eV (2)A1.6 * 1019 CB(100 V) v = = = 5.9 * 106 ms. B m C A9.1 * 1031 kgB Then h A6.63 * 1034 JsB l = = = 1.2 * 10 10 m, mv A9.1 * 10 31 kgBA5.9 * 106 msB or 0.12 nm.

EXERCISE E As a particle travels faster, does its de Broglie wavelength decrease,

increase, or remain the same? FIGURE 2712 Diffraction pattern EXERCISE F Return to the Chapter-Opening Question, page 771, and answer it again of electrons scattered from now. Try to explain why you may have answered differently the first time. aluminum foil, as recorded on film.Electron DiffractionFrom Example 2711, we see that electrons can have wavelengths on the orderof 1010 m, and even smaller. Although small, this wavelength can be detected:the spacing of atoms in a crystal is on the order of 1010 m and the orderly arrayof atoms in a crystal could be used as a type of diffraction grating, as was doneearlier for X-rays (see Section 2511). C. J. Davisson and L. H. Germer per-formed the crucial experiment: they scattered electrons from the surface of a metalcrystal and, in early 1927, observed that the electrons were scattered into a patternof regular peaks. When they interpreted these peaks as a diffraction pattern, thewavelength of the diffracted electron wave was found to be just that predicted byde Broglie, Eq. 278. In the same year, G. P. Thomson (son of J. J. Thomson) useda different experimental arrangement and also detected diffraction of electrons.(See Fig. 2712. Compare it to X-ray diffraction, Section 2511.) Later experimentsshowed that protons, neutrons, and other particles also have wave properties.

SECTION 278 Wave Nature of Matter 783

Thus the waveparticle duality applies to material objects as well as to light. The principle of complementarity applies to matter as well. That is, we must be aware of both the particle and wave aspects in order to have an understanding of matter, including electrons. But again we must recognize that a visual picture of a waveparticle is not possible.

PHYSICS APPLIED EXAMPLE 27;12 Electron diffraction. The wave nature of electrons is mani- Electron diffraction fested in experiments where an electron beam interacts with the atoms on the surface of a solid, especially crystals. By studying the angular distribution of the diffracted electrons, one can indirectly measure the geometrical arrangement of beam electron Incident

atoms. Assume that the electrons strike perpendicular to the surface of a solid (see Fig. 2713), and that their energy is low, ke = 100 eV, so that they interact only with the surface layer of atoms. If the smallest angle at which a diffraction u maximum occurs is at 24, what is the separation d between the atoms on the d sin u surface? u SOLUTION Treating the electrons as waves, we need to determine the condi- tion where the difference in path traveled by the wave diffracted from adjacent d atoms is an integer multiple of the de Broglie wavelength, so that constructive interference occurs. The path length difference is d sin u (Fig. 2713); so for the smallest value of u we must haveFIGURE 2713 Example 2712. d sin u = l.The red dots represent atoms in anorderly array in a solid. However, l is related to the (non-relativistic) kinetic energy ke by p2 h2 . ke = = 2me 2me l2 Thus h l = 32me ke A6.63 * 1034 JsB = = 0.123 nm. 31 19 32A9.11 * 10 kgB(100 eV)A1.6 * 10 JeVB The surface inter-atomic spacing is l 0.123 nm d = = = 0.30 nm. sin u sin 24 NOTE Experiments of this type verify both the wave nature of electrons and the orderly array of atoms in crystalline solids.

What Is an Electron? We might ask ourselves: What is an electron? The early experiments of J. J. Thomson (Section 271) indicated a glow in a tube, and that glow moved when a magnetic field was applied. The results of these and other experiments were best interpreted as being caused by tiny negatively charged particles which we now call electrons. No one, however, has actually seen an electron directly. The drawings we sometimes make of electrons as tiny spheres with a negative charge on them are merely convenient pictures (now recognized to be inaccurate). Again we must rely on experimental results, some of which are best interpreted using the particle model and others using the wave model. These models are mere pictures that we use to extrapolate from the macroscopic world to the tiny microscopic world of the atom. And there is no reason to expect that these models somehow reflect the reality of an electron. We thus use a wave or a particle model (whichever works best in a situation) so that we can talk about what is happening. But we should not be led to believe that an electron is a wave or a particle. Instead we could say that an electron is the set of its properties that we can measure. Bertrand Russell said it well when he wrote that an electron is a logical construction.

784 CHAPTER 27 Early Quantum Theory and Models of the Atom

279 Electron MicroscopesThe idea that electrons have wave properties led to the development of the PHYSICS APPLIEDelectron microscope (EM), which can produce images of much greater magnifi- Electron microscopecation than a light microscope. Figures 2714 and 2715 are diagrams of two types,developed around the middle of the twentieth century: the transmission electronmicroscope (TEM), which produces a two-dimensional image, and the scanningelectron microscope (SEM), which produces images with a three-dimensional quality. Electron source

Hot filament (source of electrons)

Magnetic lens High voltage +

+ + Condensing Electronics lens and screen FIGURE 2714 Transmission electron Scanning Specimen microscope. The magnetic field coils coils Electron are designed to be magnetic lenses, collector which bend the electron paths and Objective bring them to a focus, as shown. The lens sensors of the image measure electron Specimen Secondary intensity only, no color. electrons Projection lens FIGURE 2715 Scanning electron (eyepiece) microscope. Scanning coils move an electron beam back and forth across the specimen. Secondary electrons produced when the beam strikes the Image (on screen, film, specimen are collected and their or semiconductor detector) intensity affects the brightness of pixels in a monitor to produce aIn both types, the objective and eyepiece lenses are actually magnetic fields that picture.exert forces on the electrons to bring them to a focus. The fields are produced bycarefully designed current-carrying coils of wire. Photographs using each type areshown in Fig. 2716. EMs measure the intensity of electrons, producing mono-chromatic photos. Color is often added artificially to highlight.

(a) (b) (c)

FIGURE 2716 Electron micrographs,

As discussed in Sections 257 and 258, the maximum resolution of details in false color, of (a) viruses attackingon an object is about the size of the wavelength of the radiation used to view it. a cell of the bacterium EscherichiaElectrons accelerated by voltages on the order of 105 V have wavelengths of coli (TEM, L 50,000 *). (b) Sameabout 0.004 nm. The maximum resolution obtainable would be on this order, but subject by an SEM (L 35,000*).in practice, aberrations in the magnetic lenses limit the resolution in transmission (c) SEM image of an eyes retinaelectron microscopes to about 0.1 to 0.5 nm. This is still 1000 times better than a (Section 252); the rods and conesvisible-light microscope, and corresponds to a useful magnification of about a have been colored beige and green,million. Such magnifications are difficult to achieve, and more common magnifi- respectively. Part (c) is also on the cover of this book.cations are 104 to 105. The maximum resolution of a scanning electron microscopeis less, typically 5 to 10 nm although new high-resolution SEMs approach 1 nm. SECTION 279 785 PHYSICS APPLIED The scanning tunneling electron microscope (STM), developed in the 1980s, STM and AFM contains a tiny probe, whose tip may be only one (or a few) atoms wide, that is moved across the specimen to be examined in a series of linear passes. The tip, as Scanning it scans, remains very close to the surface of the specimen, about 1 nm above it, probe Fig. 2717. A small voltage applied between the probe and the surface causes electrons to leave the surface and pass through the vacuum to the probe, by a process known as tunneling (discussed in Section 3012). This tunneling Electron tunneling current is very sensitive to the gap width, so a feedback mechanism can be used Vacuum current to raise and lower the probe to maintain a constant electron current. The probes vertical motion, following the surface of the specimen, is then plotted as a function of position, scan after scan, producing a three-dimensional image of the Surface of specimen surface. Surface features as fine as the size of an atom can be resolved: a resolu- tion better than 50 pm (0.05 nm) laterally and 0.01 to 0.001 nm vertically. ThisFIGURE 2717 The probe tip of a kind of resolution has given a great impetus to the study of the surface structurescanning tunneling electron of materials. The topographic image of a surface actually represents themicroscope, as it is movedhorizontally, automatically moves up distribution of electron charge.and down to maintain a constant The atomic force microscope (AFM), developed in the 1980s, is in manytunneling current, and this motion is ways similar to an STM, but can be used on a wider range of sample materials.translated into an image of the Instead of detecting an electric current, the AFM measures the force between asurface. cantilevered tip and the sample, a force which depends strongly on the tipsample separation at each point. The tip is moved as for the STM.

FIGURE 2718 Plum-pudding 2710 Early Models of the Atom

model of the atom. The idea that matter is made up of atoms was accepted by most scientists by 1900. 1010 m With the discovery of the electron in the 1890s, scientists began to think of the atom itself as having a structure with electrons as part of that structure. We now discuss how our modern view of the atom developed, and the quantum theory with which it is intertwined. A typical model of the atom in the 1890s visualized the atom as a homogene-

ous sphere of positive charge inside of which there were tiny negatively charged electrons, a little like plums in a pudding, Fig. 2718. Around 1911, Ernest Rutherford (18711937) and his colleagues performed Positively experiments whose results contradicted the plum-pudding model of the atom. In charged these experiments a beam of positively charged alpha (a) particles was directed material at a thin sheet of metal foil such as gold, Fig. 2719. (These newly discovered a particles were emitted by certain radioactive materials and were soon shown to be doubly ionized helium atomsthat is, having a charge of 2e. ) It was Viewing screen part FIGURE 2719 Experimental setup icles for Rutherfords experiment: a particles emitted by radon are Source containing Metal deflected by the atoms of a thin metal radon foil foil and a few rebound backward.

expected from the plum-pudding model that the alpha particles would not be deflected significantly because electrons are so much lighter than alpha particles, and the alpha particles should not have encountered any massive concentration of positive charge to strongly repel them. The experimental results completely contradicted these predictions. It was found that most of the alpha particles passed through the foil unaffected, as if the foil were mostly empty space.

Some readers may say: Tell us the facts as we know them today, and dont bother us with the histor- ical background and its outmoded theories. Such an approach would ignore the creative aspect of science and thus give a false impression of how science develops. Moreover, it is not really possible to understand todays view of the atom without insight into the concepts that led to it.

786 CHAPTER 27 Early Quantum Theory and Models of the Atom

And of those deflected, a few were deflected at very large anglessome even Nucleusbackward, nearly in the direction from which they had come. This could happen, + + particleRutherford reasoned, only if the positively charged alpha particles were beingrepelled by a massive positive charge concentrated in a very small region of space FIGURE 2720 Backward rebound(see Fig. 2720). He hypothesized that the atom must consist of a tiny but mas- of a particles in Fig. 2719 explainedsive positively charged nucleus, containing over 99.9% of the mass of the atom, as the repulsion from a heavysurrounded by much lighter electrons some distance away. The electrons would positively charged nucleus.be moving in orbits about the nucleusmuch as the planets move around the FIGURE 2721 Rutherfords modelSunbecause if they were at rest, they would fall into the nucleus due to electri- of the atom: electrons orbit a tinycal attraction. See Fig. 2721. Rutherfords experiments suggested that the nucleus positive nucleus (not to scale). The atommust have a radius of about 1015 to 1014 m. From kinetic theory, and especially is visualized as mostly empty space.Einsteins analysis of Brownian motion (see Section 131), the radius of atomswas estimated to be about 1010 m. Thus the electrons would seem to be at a

distance from the nucleus of about 10,000 to 100,000 times the radius of thenucleus itself. (If the nucleus were the size of a baseball, the atom would have the +diameter of a big city several kilometers across.) So an atom would be mostlyempty space. 10 15 m Rutherfords planetary model of the atom (also called the nuclear modelof the atom) was a major step toward how we view the atom today. It was not, 10 10 mhowever, a complete model and presented some major problems, as we shall see. FIGURE 2722 Gas-discharge tube:

2711 Atomic Spectra: Key to the (a) diagram; (b) photo of an actual discharge tube for hydrogen. Structure of the Atom + AnodeEarlier in this Chapter we saw that heated solids (as well as liquids and dense gases)emit light with a continuous spectrum of wavelengths. This radiation is assumed tobe due to oscillations of atoms and molecules, which are largely governed by the interaction of each atom or molecule with its neighbors. + High Rarefied gases can also be excited to emit light. This is done by intense voltageheating, or more commonly by applying a high voltage to a discharge tubecontaining the gas at low pressure, Fig. 2722. The radiation from excited gases had been observed early in the nineteenth century, and it was found that the spectrum was not continuous. Rather, excited gases emit light of only certainwavelengths, and when this light is analyzed through the slit of a spectroscope Cathode (a)or spectrometer, a line spectrum is seen rather than a continuous spectrum.The line spectra emitted by a number of elements in the visible region are shownbelow in Fig. 2723, and in Chapter 24, Fig. 2428. The emission spectrum ischaracteristic of the material and can serve as a type of fingerprint foridentification of the gas. We also saw (Chapter 24) that if a continuous spectrum passes through ararefied gas, dark lines are observed in the emerging spectrum, at wavelengthscorresponding to lines normally emitted by the gas. This is called an absorptionspectrum (Fig. 2723c), and it became clear that gases can absorb light at thesame frequencies at which they emit. Using film sensitive to ultraviolet and toinfrared light, it was found that gases emit and absorb discrete frequencies inthese regions as well as in the visible. (b)

(a)

(b) FIGURE 2723 Emission spectra of

the gases (a) atomic hydrogen, (b) helium, and (c) the solar absorption spectrum.(c) SECTION 2711 Atomic Spectra: Key to the Structure of the Atom 787 365 In low-density gases, the atoms are far apart on average and hence the light emitted or absorbed is assumed to be by individual atoms rather than through UV interactions between atoms, as in a solid, liquid, or dense gas. Thus the line spectra serve as a key to the structure of the atom: any theory of atomic structure 410 Violet must be able to explain why atoms emit light only of discrete wavelengths, and it should be able to predict what these wavelengths are. 434 Blue Hydrogen is the simplest atomit has only one electron orbiting its nucleus. It also has the simplest spectrum. The spectrum of most atoms shows little apparent regularity. But the spacing between lines in the hydrogen spectrum Blue- decreases in a regular way, Fig. 2724. Indeed, in 1885, J. J. Balmer (18251898) 486 green showed that the four lines in the visible portion of the hydrogen spectrum (with measured wavelengths 656 nm, 486 nm, 434 nm, and 410 nm) have wavelengths that fit the formula

(nm) 1 1 1 = R 2 - 2 , n = 3, 4, p . (27;9) l 2 n Here n takes on the values 3, 4, 5, 6 for the four visible lines, and R, called the Rydberg constant, has the value R = 1.0974 * 107 m1. Later it was found that this Balmer series of lines extended into the UV region, ending at l = 365 nm, as shown in Fig. 2724. Balmers formula, Eq. 279, also worked for these lines with higher integer values of n. The lines near 365 nm become too close together to distinguish, but the limit of the series at 365 nm corresponds to n = q (so 1n2 = 0 in Eq. 279). 656 Red Later experiments on hydrogen showed that there were similar series of linesFIGURE 2724 Balmer series of in the UV and IR regions, and each series had a pattern just like the Balmerlines for hydrogen. series, but at different wavelengths, Fig. 2725. Each of these series was found to

FIGURE 2725 Line spectrum of Wavelength,

atomic hydrogen. Each series fits the

1875 nm 122 nm

365 nm

656 nm

820 nm 91 nm

1 1 1formula = R 2 - 2 where l n nn = 1 for the Lyman series,n = 2 for the Balmer series,n = 3 for the Paschen series,and so on; n can take on all integervalues from n = n + 1 up to infinity. Lyman Balmer series Paschen seriesThe only lines in the visible region seriesof the electromagnetic spectrum arepart of the Balmer series. UV Visible light IR

fit a formula with the same form as Eq. 279 but with the 12 2 replaced by 112, 132, 14 2, and so on. For example, the Lyman series contains lines with wavelengths from 91 nm to 122 nm (in the UV region) and fits the formula

1 1 1 = R 2 - 2 , n = 4, 5, p . l 3 n The Rutherford model was unable to explain why atoms emit line spectra. It had other difficulties as well. According to the Rutherford model, electrons orbit the nucleus, and since their paths are curved the electrons are accelerating. Hence they should give off light like any other accelerating electric charge (Chapter 22).

788 CHAPTER 27 Early Quantum Theory and Models of the Atom

Since light carries off energy and energy is conserved, the electrons own energymust decrease to compensate. Hence electrons would be expected to spiral into thenucleus. As they spiraled inward, their frequency would increase in a short timeand so too would the frequency of the light emitted. Thus the two main difficultiesof the Rutherford model are these: (1) it predicts that light of a continuous range offrequencies will be emitted, whereas experiment shows line spectra; (2) it predictsthat atoms are unstableelectrons would quickly spiral into the nucleusbut weknow that atoms in general are stable, because there is stable matter all around us. Clearly Rutherfords model was not sufficient. Some sort of modificationwas needed, and Niels Bohr provided it in a model that included the quantumhypothesis. Although the Bohr model has been superseded, it did provide acrucial stepping stone to our present understanding. And some aspects of theBohr model are still useful today, so we examine it in detail in the next Section.

2712 The Bohr Model

Bohr had studied in Rutherfords laboratory for several months in 1912 and wasconvinced that Rutherfords planetary model of the atom had validity. But inorder to make it work, he felt that the newly developing quantum theory wouldsomehow have to be incorporated in it. The work of Planck and Einstein hadshown that in heated solids, the energy of oscillating electric charges must changediscontinuouslyfrom one discrete energy state to another, with the emission ofa quantum of light. Perhaps, Bohr argued, the electrons in an atom also cannotlose energy continuously, but must do so in quantum jumps. In working out hismodel during the next year, Bohr postulated that electrons move about the nucleusin circular orbits, but that only certain orbits are allowed. He further postulatedthat an electron in each orbit would have a definite energy and would move in the Euorbit without radiating energy (even though this violated classical ideas sinceaccelerating electric charges are supposed to emit EM waves; see Chapter 22). hfHe thus called the possible orbits stationary states. In this Bohr model, light is Elemitted only when an electron jumps from a higher (upper) stationary state toanother of lower energy, Fig. 2726. When such a transition occurs, a single pho- FIGURE 2726 An atom emits aton of light is emitted whose energy, by energy conservation, is given by photon (energy = hf) when its energy changes from Eu to a lower hf = Eu - El , (27;10) energy El .where Eu refers to the energy of the upper state and El the energy of the lowerstate. In 191213, Bohr set out to determine what energies these orbits would havein the simplest atom, hydrogen; the spectrum of light emitted could then be pre-dicted from Eq. 2710. In the Balmer formula he had the key he was looking for.Bohr quickly found that his theory would agree with the Balmer formula ifhe assumed that the electrons angular momentum L is quantized and equal to aninteger n times h2p. As we saw in Chapter 8 angular momentum is given byL = Iv, where I is the moment of inertia and v is the angular velocity. For asingle particle of mass m moving in a circle of radius r with speed v, I = mr2 andv = vr; hence, L = Iv = Amr2 B(vr) = mvr. Bohrs quantum condition is h , L = mvrn = n n = 1, 2, 3, p , (27;11) 2pwhere n is an integer and rn is the radius of the nth possible orbit. The allowedorbits are numbered 1, 2, 3, p , according to the value of n, which is called theprincipal quantum number of the orbit. Equation 2711 did not have a firm theoretical foundation. Bohr had searchedfor some quantum condition, and such tries as E = hf (where E representsthe energy of the electron in an orbit) did not give results in accord with experi-ment. Bohrs reason for using Eq. 2711 was simply that it worked; and we nowlook at how. In particular, let us determine what the Bohr theory predicts for themeasurable wavelengths of emitted light.

We include Z in our derivation so that we can treat other single-electron (hydrogenlike) atoms r4 16r1 such as the ions He (Z = 2) and Li2 (Z = 3). Helium in the neutral state has two electrons; if one electron is missing, the remaining He ion consists of one electron revolving around a nucleus of charge 2e. Similarly, doubly ionized lithium, Li2, also has a single electron, and in this case Z = 3.

Be careful not to believe that these well-defined orbits actually exist. Today electrons are better thought of as forming clouds, as discussed in Chapter 28.790 CHAPTER 27 In each of its possible orbits, the electron in a Bohr model atom would have adefinite energy, as the following calculation shows. The total energy equals thesum of the kinetic and potential energies. The potential energy of the electron isgiven by pe = qV = eV, where V is the potential due to a point charge Ze asgiven by Eq. 175: V = kQr = kZer. So Ze2 . pe = eV = k rThe total energy En for an electron in the nth orbit of radius rn is the sum of thekinetic and potential energies: kZe2 . En = 2 mv 1 2 - rnWhen we substitute v from Eq. 2711 and rn from Eq. 2712 into this equation,we obtain 2p2Z2e4mk2 1 En = n = 1, 2, 3, p . (27;15a) h2 n2If we evaluate the constant term in Eq. 2715a and convert it to electron volts, asis customary in atomic physics, we obtain Z2 , En = (13.6 eV) n = 1, 2, 3, p . (27;15b) n2The lowest energy level (n = 1) for hydrogen (Z = 1) is E1 = 13.6 eV.Since n appears in the denominator of Eq. 2715b, the energies of the larger 2

orbits in hydrogen (Z = 1) are given by

13.6 eV . En = n2For example, 13.6 eV E2 = = 3.40 eV, 4 13.6 eV E3 = = 1.51 eV. 9We see that not only are the orbit radii quantized, but from Eqs. 2715, so is theenergy. The quantum number n that labels the orbit radii also labels the energylevels. The lowest energy level or energy state has energy E1 , and is called theground state. The higher states, E2 , E3 , and so on, are called excited states. Thefixed energy levels are also called stationary states. Notice that although the energy for the larger orbits has a smaller numericalvalue, all the energies are less than zero. Thus, 3.4 eV is a higher energy than13.6 eV. Hence the orbit closest to the nucleus Ar1 B has the lowest energy (themost negative). The reason the energies have negative values has to do with the waywe defined the zero for potential energy. For two point charges, pe = kq1 q2rcorresponds to zero potential energy when the two charges are infinitely far apart(Section 175). Thus, an electron that can just barely be free from the atom byreaching r = q (or, at least, far from the nucleus) with zero kinetic energy willhave E = ke + pe = 0 + 0 = 0, corresponding to n = q in Eqs. 2715. If anelectron is free and has kinetic energy, then E 7 0. To remove an electron that ispart of an atom requires an energy input (otherwise atoms would not be stable).Since E 0 for a free electron, then an electron bound to an atom needs to haveE 6 0. That is, energy must be added to bring its energy up, from a negativevalue to at least zero in order to free it. The minimum energy required to remove an electron from an atom initiallyin the ground state is called the binding energy or ionization energy. The ionizationenergy for hydrogen has been measured to be 13.6 eV, and this correspondsprecisely to removing an electron from the lowest state, E1 = 13.6 eV, up toE = 0 where it can be free.

SECTION 2712 The Bohr Model 791

Spectra Lines Explained It is useful to show the various possible energy values as horizontal lines on an energy-level diagram. This is shown for hydrogen in Fig. 2729. The electron in a hydrogen atom can be in any one of these levels according to Bohr theory. But it could never be in between, say at 9.0 eV. At room temperature, nearly all H atoms will be in the ground state (n = 1). At higher temperatures, or during an electric discharge when there are many collisions between free electrons and atoms, many atoms can be in excited states (n 7 1). Once in an excited state, an atoms electron can jump down to a lower state, and give off a photon in the process. This is, according to the Bohr model, the origin of the emission spectra of excited gases. Note that above E = 0, an electron is free and can have any energy (E is not quantized). Thus there is a continuum of energy states above E = 0, as indi- cated in the energy-level diagram of Fig. 2729.

transitions for the spectral lines of the

Lyman, Balmer, and Paschen series (Fig. 2725). Each vertical arrow represents an atomic transition that gives rise to the photons of one spectral line (a single wavelength or frequency). 10

n=1 Ground state

13.6 Lyman 15 series

The vertical arrows in Fig. 2729 represent the transitions or jumps that correspond to the various observed spectral lines. For example, an electron jumping from the level n = 3 to n = 2 would give rise to the 656-nm line in the Balmer series, and the jump from n = 4 to n = 2 would give rise to the 486-nm line (see Fig. 2724). We can predict wavelengths of the spectral lines emitted according to Bohr theory by combining Eq. 2710 with Eq. 2715. Since hf = hcl, we have from Eq. 2710 1 hf 1 = = AE - En B, l hc hc n where n refers to the upper state and n to the lower state. Then using Eq. 2715, 1 2p2Z2e4mk2 1 1 = 2 - 2 . (27;16) l hc 3 n n This theoretical formula has the same form as the experimental Balmer formula, Eq. 279, with n = 2. Thus we see that the Balmer series of lines corresponds to transitions or jumps that bring the electron down to the second energy level. Similarly, n = 1 corresponds to the Lyman series and n = 3 to the Paschen series (see Fig. 2729). When the constant in Eq. 2716 is evaluated with Z = 1, it is found to have the measured value of the Rydberg constant, R = 1.0974 * 107 m1 in Eq. 279, in accord with experiment (see Problem 54).792 CHAPTER 27 The great success of Bohrs model is that it gives an explanation for why atomsemit line spectra, and accurately predicts the wavelengths of emitted light forhydrogen. The Bohr model also explains absorption spectra: photons of just theright wavelength can knock an electron from one energy level to a higher one. Toconserve energy, only photons that have just the right energy will be absorbed. Thisexplains why a continuous spectrum of light entering a gas will emerge with dark(absorption) lines at frequencies that correspond to emission lines (Fig. 2723c). The Bohr theory also ensures the stability of atoms. It establishes stability bydecree: the ground state is the lowest state for an electron and there is no lowerenergy level to which it can go and emit more energy. Finally, as we saw above,the Bohr theory accurately predicts the ionization energy of 13.6 eV for hydro-gen. However, the Bohr model was not so successful for other atoms, and hasbeen superseded as we shall discuss in the next Chapter. We discuss the Bohrmodel because it was an important start and because we still use the concept ofstationary states, the ground state, and transitions between states. Also, the ter-minology used in the Bohr model is still used by chemists and spectroscopists. EXAMPLE 27;13 Wavelength of a Lyman line. Use Fig. 2729 to deter- mine the wavelength of the first Lyman line, the transition from n = 2 to n = 1. In what region of the electromagnetic spectrum does this lie? APPROACH We use Eq. 2710, hf = Eu - El , with the energies obtained from Fig. 2729 to find the energy and the wavelength of the transition. The region of the electromagnetic spectrum is found using the EM spectrum in Fig. 228. SOLUTION In this case, hf = E2 - E1 = E3.4 eV - (13.6 eV)F = 10.2 eV = (10.2 eV)A1.60 * 1019 JeVB = 1.63 * 10 18 J. Since l = cf, we have c hc A6.63 * 10 34 JsBA3.00 * 108 msB l = = = = 1.22 * 10 7 m, f E2 - E1 1.63 * 1018 J or 122 nm, which is in the UV region of the EM spectrum, Fig. 228. See also Fig. 2725, where this value is confirmed experimentally. NOTE An alternate approach: use Eq. 2716 to find l, and get the same result.

EXAMPLE 27;15 Absorption wavelength. Use Fig. 2729 to determine the

maximum wavelength that hydrogen in its ground state can absorb. What would be the next smaller wavelength that would work? APPROACH Maximum wavelength corresponds to minimum energy, and this would be the jump from the ground state up to the first excited state (Fig. 2729). The next smaller wavelength occurs for the jump from the ground state to the second excited state. SOLUTION The energy needed to jump from the ground state to the first excited state is 13.6 eV - 3.4 eV = 10.2 eV; the required wavelength, as we saw in Example 2713, is 122 nm. The energy to jump from the ground state to the second excited state is 13.6 eV - 1.5 eV = 12.1 eV, which corresponds to a wavelength c hc hc A6.63 * 1034 JsBA3.00 * 108 msB l = = = = = 103 nm. f hf E3 - E1 (12.1 eV)A1.60 * 1019 JeVB SECTION 2712 793 EXAMPLE 27;16 He ionization energy. (a) Use the Bohr model to deter- mine the ionization energy of the He ion, which has a single electron. (b) Also calculate the maximum wavelength a photon can have to cause ionization. The helium atom is the second atom, after hydrogen, in the Periodic Table (next Chapter); its nucleus contains 2 protons and normally has 2 electrons circulating around it, so Z = 2. APPROACH We want to determine the minimum energy required to lift the electron from its ground state and to barely reach the free state at E = 0. The ground state energy of He is given by Eq. 2715b with n = 1 and Z = 2. SOLUTION (a) Since all the symbols in Eq. 2715b are the same as for the calculation for hydrogen, except that Z is 2 instead of 1, we see that E1 will be Z2 = 2 2 = 4 times the E1 for hydrogen: E1 = 4(13.6 eV) = 54.4 eV. Thus, to ionize the He ion should require 54.4 eV, and this value agrees with experiment. (b) The maximum wavelength photon that can cause ionization will have energy hf = 54.4 eV and wavelength c hc A6.63 * 10 34 JsBA3.00 * 108 msB l = = = = 22.8 nm. f hf (54.4 eV)A1.60 * 1019 JeVB If l 7 22.8 nm, ionization can not occur. NOTE If the atom absorbed a photon of greater energy (wavelength shorter than 22.8 nm), the atom could still be ionized and the freed electron would have kinetic energy of its own.

In this Example 2716, we saw that E1 for the He ion is four times more negative than that for hydrogen. Indeed, the energy-level diagram for He looks just like that for hydrogen, Fig. 2729, except that the numerical values for each energy level are four times larger. Note, however, that we are talking here about the He ion. Normal (neutral) helium has two electrons and its energy level diagram is entirely different. CONCEPTUAL EXAMPLE 27;17 Hydrogen at 20C. (a) Estimate the average kinetic energy of whole hydrogen atoms (not just the electrons) at room temperature. (b) Use the result to explain why, at room temperature, very few H atoms are in excited states and nearly all are in the ground state, and hence emit no light. RESPONSE According to kinetic theory (Chapter 13), the average kinetic energy of atoms or molecules in a gas is given by Eq. 138: G = 32 kT, where k = 1.38 * 10 23 JK is Boltzmanns constant, and T is the kelvin (absolute) temperature. Room temperature is about T = 300 K, so G = 32 A1.38 * 10 23 JKB(300 K) = 6.2 * 10 21 J, or, in electron volts: 6.2 * 1021 J G = = 0.04 eV. 1.6 * 10 19 JeV The average ke of an atom as a whole is thus very small compared to the energy between the ground state and the next higher energy state (13.6 eV - 3.4 eV = 10.2 eV). Any atoms in excited states quickly fall to the ground state and emit light. Once in the ground state, collisions with other atoms can transfer energy of only 0.04 eV on the average. A small fraction of atoms can have much more energy (see Section 1310 on the distribution of molecular speeds), but even a kinetic energy that is 10 times the average is not nearly enough to excite atoms into states above the ground state. Thus, at room temperature, practically all atoms are in the ground state. Atoms can be excited to upper states by very high temperatures, or by applying a high voltage so a current of high energy electrons passes through the gas as in a discharge tube (Fig. 2722).794 CHAPTER 27Correspondence PrincipleWe should note that Bohr made some radical assumptions that were at variancewith classical ideas. He assumed that electrons in fixed orbits do not radiate lighteven though they are accelerating (moving in a circle), and he assumed thatangular momentum is quantized. Furthermore, he was not able to say how anelectron moved when it made a transition from one energy level to another. Onthe other hand, there is no real reason to expect that in the tiny world of the atomelectrons would behave as ordinary-sized objects do. Nonetheless, he felt thatwhere quantum theory overlaps with the macroscopic world, it should predictclassical results. This is the correspondence principle, already mentioned inregard to relativity (Section 2611). This principle does work for Bohrs theoryof the hydrogen atom. The orbit sizes and energies are quite different for n = 1and n = 2, say. But orbits with n = 100,000,000 and 100,000,001 would be veryclose in radius and energy (see Fig. 2729). Indeed, transitions between suchlarge orbits, which would approach macroscopic sizes, would be imperceptible.Such orbits would thus appear to be continuously spaced, which is what weexpect in the everyday world. Finally, it must be emphasized that the well-defined orbits of the Bohr modeldo not actually exist. The Bohr model is only a model, not reality. The idea ofelectron orbits was rejected a few years later, and today electrons are thought of(Chapter 28) as forming probability clouds. (a)2713 de Broglies Hypothesis Applied to AtomsBohrs theory was largely of an ad hoc nature. Assumptions were made so thattheory would agree with experiment. But Bohr could give no reason why the orbitswere quantized, nor why there should be a stable ground state. Finally, ten yearslater, a reason was proposed by Louis de Broglie. We saw in Section 278 thatin 1923, de Broglie proposed that material particles, such as electrons, have a wave (b)nature; and that this hypothesis was confirmed by experiment several years later. One of de Broglies original arguments in favor of the wave nature of elec- FIGURE 2730 (a) An ordinarytrons was that it provided an explanation for Bohrs theory of the hydrogen standing wave compared to aatom. According to de Broglie, a particle of mass m moving with a nonrelativistic circular standing wave. (b) When aspeed v would have a wavelength (Eq. 278) of wave does not close (and hence interferes destructively with itself), h . l = it rapidly dies out. mvEach electron orbit in an atom, he proposed, is actually a standing wave. As we FIGURE 2731 Standing circularsaw in Chapter 11, when a violin or guitar string is plucked, a vast number of waves for two, three, and fivewavelengths are excited. But only certain onesthose that have nodes at the wavelengths on the circumference;endsare sustained. These are the resonant modes of the string. Waves with n, the number of wavelengths, isother wavelengths interfere with themselves upon reflection and their amplitudes also the quantum number.quickly drop to zero. With electrons moving in circles, according to Bohrs theory,de Broglie argued that the electron wave was a circular standing wave that closeson itself, Fig. 2730a. If the wavelength of a wave does not close on itself, as inFig. 2730b, destructive interference takes place as the wave travels around theloop, and the wave quickly dies out. Thus, the only waves that persist are those forwhich the circumference of the circular orbit contains a whole number of wave-lengths, Fig. 2731. The circumference of a Bohr orbit of radius rn is 2prn , so to n2have constructive interference, we need 2prn = nl, n = 1, 2, 3, p .When we substitute l = hmv, we get 2prn = nhmv, or n3 nh . mvrn = 2pThis is just the quantum condition proposed by Bohr on an ad hoc basis, Eq. 2711.It is from this equation that the discrete orbits and energy levels were derived. n5

SECTION 2713 de Broglies Hypothesis Applied to Atoms 795

Now we have a first explanation for the quantized orbits and energy states in the Bohr model: they are due to the wave nature of the electron, and only resonant standing waves can persist. This implies that the waveparticle duality is at the root of atomic structure. In viewing the circular electron waves of Fig. 2731, the electron is not to be thought of as following the oscillating wave pattern. In the Bohr model of hydrogen, n2 the electron moves in a circle. The circular wave, on the other hand, represents the amplitude of the electron matter wave, and in Fig. 2731 the wave amplitude is shown superimposed on the circular path of the particle orbit for convenience. n3 Bohrs theory worked well for hydrogen and for one-electron ions. But it did not prove successful for multi-electron atoms. Bohr theory could not predict line spectra even for the next simplest atom, helium. It could not explain why some emis- sion lines are brighter than others, nor why some lines are split into two or more closely spaced lines (fine structure). A new theory was needed and was indeed n5 developed in the 1920s. This new and radical theory is called quantum mechanics.FIGURE 2731 (Repeated.) Standing It finally solved the problem of atomic structure, but it gives us a very differentcircular waves for two, three, and view of the atom: the idea of electrons in well-defined orbits was replaced with thefive wavelengths on the circumference;n, the number of wavelengths, idea of electron clouds. This new theory of quantum mechanics has given us ais also the quantum number. wholly different view of the basic mechanisms underlying physical processes.

We note, however, that Eq. 2711 is no longer considered valid, as discussed in the next Chapter.

SummaryThe electron was discovered using an evacuated cathode ray The principle of complementarity states that we must betube. The measurement of the charge-to-mass ratio (em) of aware of both the particle and wave properties of light and ofthe electron was done using magnetic and electric fields. The matter for a complete understanding of them.charge e on the electron was first measured in the Millikan Electron microscopes (EM) make use of the wave proper-oil-drop experiment and then its mass was obtained from the ties of electrons to form an image: their lenses are magnetic.measured value of the em ratio. Various types of EM exist: some can magnify 100,000* (1000* Quantum theory has its origins in Plancks quantum hypothesis better than a light microscope); others can give a 3-D image.that molecular oscillations are quantized: their energy E can only Early models of the atom include Rutherfords planetarybe integer (n) multiples of hf, where h is Plancks constant and (or nuclear) model of an atom which consists of a tiny butf is the natural frequency of oscillation: massive positively charged nucleus surrounded (at a relatively E = nhf. (27;3) great distance) by electrons. To explain the line spectra emitted by atoms, as well as theThis hypothesis explained the spectrum of radiation emitted by stability of atoms, the Bohr model postulated that: (1) elec-a blackbody at high temperature. trons bound in an atom can only occupy orbits for which the Einstein proposed that for some experiments, light could angular momentum is quantized, which results in discretebe pictured as being emitted and absorbed as quanta (particles), values for the radius and energy; (2) an electron in such awhich we now call photons, each with energy stationary state emits no radiation; (3) if an electron jumps to E = hf (27;4) a lower state, it emits a photon whose energy equals theand momentum difference in energy between the two states; (4) the angular E hf h. momentum L of atomic electrons is quantized by the rule p = = = (27;6) L = nh2p, where n is an integer called the quantum number. c c l The n = 1 state is the ground state, which in hydrogen hasHe proposed the photoelectric effect as a test for the photon an energy E1 = 13.6 eV. Higher values of n correspond totheory of light. In the photoelectric effect, the photon theory excited states, and their energies aresays that each incident photon can strike an electron in amaterial and eject it if the photon has sufficient energy. The Z2 ,maximum energy of ejected electrons is then linearly related to En = (13.6 eV) (27;15b) n2the frequency of the incident light. The photon theory is also supported by the Compton where Ze is the charge on the nucleus. Atoms are excited toeffect and the observation of electronpositron pair production. these higher states by collisions with other atoms or electrons, The wave;particle duality refers to the idea that light and or by absorption of a photon of just the right frequency.matter (such as electrons) have both wave and particle proper- De Broglies hypothesis that electrons (and other matter)ties. The wavelength of an object is given by have a wavelength l = hmv gave an explanation for Bohrs quantized orbits by bringing in the waveparticle duality: the h, l = (27;8) orbits correspond to circular standing waves in which the cir- p cumference of the orbit equals a whole number of wavelengths.where p is the momentum of the object ( p = mv for a particleof mass m and speed v).

796 CHAPTER 27 Early Quantum Theory and Models of the Atom

Questions 1. Does a lightbulb at a temperature of 2500 K produce as 20. When a wide spectrum of light passes through hydrogen white a light as the Sun at 6000 K? Explain. gas at room temperature, absorption lines are observed 2. If energy is radiated by all objects, why can we not see them that correspond only to the Lyman series. Why dont we in the dark? (See also Section 148.) observe the other series?

3. What can be said about the relative temperatures of 21. How can you tell if there is oxygen near the surface of the whitish-yellow, reddish, and bluish stars? Explain. Sun?

4. Darkrooms for developing black-and-white film were 22. (a) List at least three successes of the Bohr model of the sometimes lit by a red bulb. Why red? Explain if such a atom, according to Section 2712. (b) List at least two bulb would work in a darkroom for developing color film. observations that the Bohr model could not explain, according to Section 2713. 5. If the threshold wavelength in the photoelectric effect increases when the emitting metal is changed to a different 23. According to Section 2711, what were the two main metal, what can you say about the work functions of the difficulties of the Rutherford model of the atom? two metals? 24. Is it possible for the de Broglie wavelength of a particle 6. Explain why the existence of a cutoff frequency in the to be greater than the dimensions of the particle? To be photoelectric effect more strongly favors a particle theory smaller? Is there any direct connection? Explain. rather than a wave theory of light. 25. How can the spectrum of hydrogen contain so many lines 7. UV light causes sunburn, whereas visible light does not. when hydrogen contains only one electron? Suggest a reason. 26. Explain how the closely spaced energy levels for hydrogen 8. The work functions for sodium and cesium are 2.28 eV and near the top of Fig. 2729 correspond to the closely spaced 2.14 eV, respectively. For incident photons of a given fre- spectral lines at the top of Fig. 2724. quency, which metal will give a higher maximum kinetic energy for the electrons? Explain. 27. In a helium atom, which contains two electrons, do you think that on average the electrons are closer to the 9. Explain how the photoelectric circuit of Fig. 276 could be nucleus or farther away than in a hydrogen atom? Why? used in (a) a burglar alarm, (b) a smoke detector, (c) a photo- graphic light meter. 28. The Lyman series is brighter than the Balmer series, 10. (a) Does a beam of infrared photons always have less because this series of transitions ends up in the most energy than a beam of ultraviolet photons? Explain. common state for hydrogen, the ground state. Why then (b) Does a single photon of infrared light always have less was the Balmer series discovered first? energy than a single photon of ultraviolet light? Why? 29. Use conservation of momentum to explain why photons 11. Light of 450-nm wavelength strikes a metal surface, and a emitted by hydrogen atoms have slightly less energy than stream of electrons emerges from the metal. If light of the that predicted by Eq. 2710. same intensity but of wavelength 400 nm strikes the sur- 30. State if a continuous or a line spectrum is produced by face, are more electrons emitted? Does the energy of the each of the following: (a) a hot solid object; (b) an emitted electrons change? Explain. excited, rarefied gas; (c) a hot liquid; (d) light from a hot*12. If an X-ray photon is scattered by an electron, does the solid that passes through a cooler rarefied gas; (e) a hot photons wavelength change? If so, does it increase or dense gas. For each, if a line spectrum is produced, is it an decrease? Explain. emission or an absorption spectrum?*13. In both the photoelectric effect and in the Compton effect, 31. Suppose we obtain an emission spectrum for hydrogen at a photon collides with an electron causing the electron to very high temperature (when some of the atoms are in fly off. What is the difference between the two processes? excited states), and an absorption spectrum at room tem- 14. Why do we say that light has wave properties? Why do we perature, when all atoms are in the ground state. Will the say that light has particle properties? two spectra contain identical lines?

15. Why do we say that electrons have wave properties? Why

do we say that electrons have particle properties? 16. What are the differences between a photon and an electron? Be specific: make a list. 17. If an electron and a proton travel at the same speed, which has the shorter wavelength? Explain. 18. An electron and a proton are accelerated through the same voltage. Which has the longer wavelength? Explain why. 19. In Rutherfords planetary model of the atom, what keeps the electrons from flying off into space?

Questions 797 MisConceptual Questions1. Which of the following statements is true regarding how 8. When you throw a baseball, its de Broglie wavelength is blackbody radiation changes as the temperature of the (a) the same size as the ball. radiating object increases? (b) about the same size as an atom. (a) Both the maximum intensity and the peak wavelength (c) about the same size as an atoms nucleus. increase. (d) much smaller than the size of an atoms nucleus. (b) The maximum intensity increases, and the peak wavelength decreases. 9. Electrons and photons of light are similar in that (c) Both the maximum intensity and the peak wavelength (a) both have momentum given by hl. decrease. (b) both exhibit waveparticle duality. (d) The maximum intensity decreases, and the peak (c) both are used in diffraction experiments to explore wavelength increases. structure. (d) All of the above.2. As red light shines on a piece of metal, no electrons are (e) None of the above. released. When the red light is slowly changed to shorter- wavelength light (basically progressing through the rainbow), 10. In Rutherfords famous set of experiments described in nothing happens until yellow light shines on the metal, at Section 2710, the fact that some alpha particles were which point electrons are released from the metal. If this deflected at large angles indicated that (choose all that apply) metal is replaced with a metal having a higher work function, (a) the nucleus was positive. which light would have the best chance of releasing elec- (b) charge was quantized. trons from the metal? (c) the nucleus was concentrated in a small region of (a) Blue. space. (b) Red. (d) most of the atom is empty space. (c) Yellow would still work fine. (e) None of the above. (d) We need to know more about the metals involved. 11. Which of the following electron transitions between two3. A beam of red light and a beam of blue light have equal energy states (n) in the hydrogen atom corresponds to the intensities. Which statement is true? emission of a photon with the longest wavelength? (a) There are more photons in the blue beam. (a) 2 S 5. (b) There are more photons in the red beam. (b) 5 S 2. (c) Both beams contain the same number of photons. (c) 5 S 8. (d) The number of photons is not related to intensity. (d) 8 S 5.4. Which of the following is necessarily true? (a) Red light has more energy than violet light. 12. If we set the potential energy of an electron and a proton (b) Violet light has more energy than red light. to be zero when they are an infinite distance apart, then (c) A single photon of red light has more energy than a the lowest energy a bound electron in a hydrogen atom can single photon of violet light. have is (d) A single photon of violet light has more energy than a (a) 0. single photon of red light. (b) 13.6 eV. (e) None of the above. (c) any possible value. (f) A combination of the above (specify). (d) any value between 13.6 eV and 0.5. If a photon of energy E ejects electrons from a metal with 13. Which of the following is the currently accepted model of kinetic energy ke, then a photon with energy E2 the atom? (a) will eject electrons with kinetic energy ke2. (a) The plum-pudding model. (b) will eject electrons with an energy greater than ke2. (b) The Rutherford atom. (c) will eject electrons with an energy less than ke2. (c) The Bohr atom. (d) might not eject any electrons. (d) None of the above.6. If the momentum of an electron were doubled, how would 14. Light has all of the following except: its wavelength change? (a) mass. (a) No change. (b) momentum. (b) It would be halved. (c) kinetic energy. (c) It would double. (d) frequency. (d) It would be quadrupled. (e) wavelength. (e) It would be reduced to one-fourth.7. Which of the following can be thought of as either a wave or a particle? (a) Light. (b) An electron. (c) A proton. (d) All of the above.

798 CHAPTER 27 Early Quantum Theory and Models of the Atom

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Problems271 Discovery of the Electron 18. (II) The human eye can respond to as little as 1018 J of light 1. (I) What is the value of em for a particle that moves in a energy. For a wavelength at the peak of visual sensitivity, circle of radius 14 mm in a 0.86-T magnetic field if a perpen- 550 nm, how many photons lead to an observable flash? dicular 640-Vm electric field will make the path straight? 19. (II) What is the longest wavelength of light that will emit 2. (II) (a) What is the velocity of a beam of electrons that go electrons from a metal whose work function is 2.90 eV? undeflected when passing through crossed (perpendicular) 20. (II) The work functions for sodium, cesium, copper, and iron electric and magnetic fields of magnitude 1.88 * 104 Vm are 2.3, 2.1, 4.7, and 4.5 eV, respectively. Which of these and 2.60 * 103 T, respectively? (b) What is the radius of metals will not emit electrons when visible light shines on it? the electron orbit if the electric field is turned off? 21. (II) In a photoelectric-effect experiment it is observed that 3. (II) An oil drop whose mass is 2.8 * 1015 kg is held at rest no current flows unless the wavelength is less than 550 nm. between two large plates separated by 1.0 cm (Fig. 273), (a) What is the work function of this material? (b) What stop- when the potential difference between the plates is 340 V. ping voltage is required if light of wavelength 400 nm is used? How many excess electrons does this drop have? 22. (II) What is the maximum kinetic energy of electrons ejected from barium AW0 = 2.48 eVB when illuminated by272 Blackbodies; Plancks Quantum Hypothesis white light, l = 400 to 750 nm? 4. (I) How hot is a metal being welded if it radiates most 23. (II) Barium has a work function of 2.48 eV. What is the strongly at 520 nm? maximum kinetic energy of electrons if the metal is illumi- 5. (I) Estimate the peak wavelength for radiation emitted from nated by UV light of wavelength 365 nm? What is their speed? (a) ice at 0C, (b) a floodlamp at 3100 K, (c) helium at 4 K, 24. (II) When UV light of wavelength 255 nm falls on a metal assuming blackbody emission. In what region of the EM surface, the maximum kinetic energy of emitted electrons spectrum is each? is 1.40 eV. What is the work function of the metal? 6. (I) (a) What is the temperature if the peak of a blackbody 25. (II) The threshold wavelength for emission of electrons spectrum is at 18.0 nm? (b) What is the wavelength at the from a given surface is 340 nm. What will be the maximum peak of a blackbody spectrum if the body is at a tempera- kinetic energy of ejected electrons when the wavelength is ture of 2200 K? changed to (a) 280 nm, (b) 360 nm? 7. (I) An HCl molecule vibrates with a natural frequency of 26. (II) A certain type of film is sensitive only to light whose 8.1 * 1013 Hz. What is the difference in energy (in joules wavelength is less than 630 nm. What is the energy (eV and and electron volts) between successive values of the oscilla- kcalmol) needed for the chemical reaction to occur which tion energy? causes the film to change? 8. (II) The steps of a flight of stairs are 20.0 cm high (vertically). 27. (II) When 250-nm light falls on a metal, the current through a If a 62.0-kg person stands with both feet on the same step, photoelectric circuit (Fig. 276) is brought to zero at a stopping what is the gravitational potential energy of this person, voltage of 1.64 V. What is the work function of the metal? relative to the ground, on (a) the first step, (b) the second 28. (II) In a photoelectric experiment using a clean sodium step, (c) the third step, (d) the nth step? (e) What is the change surface, the maximum energy of the emitted electrons was in energy as the person descends from step 6 to step 2? measured for a number of different incident frequencies, 9. (II) Estimate the peak wavelength of light emitted from the with the following results. pupil of the human eye (which approximates a blackbody) assuming normal body temperature. Frequency ( * 1014 Hz) Energy (eV)

273 and 274 Photons and the Photoelectric Effect 11.8 2.60 10.6 2.1110. (I) What is the energy of photons (joules) emitted by a 9.9 1.81 91.7-MHz FM radio station? 9.1 1.4711. (I) What is the energy range (in joules and eV) of photons 8.2 1.10 in the visible spectrum, of wavelength 400 nm to 750 nm? 6.9 0.5712. (I) A typical gamma ray emitted from a nucleus during radioactive decay may have an energy of 320 keV. What is Plot the graph of these results and find: (a) Plancks constant; its wavelength? Would we expect significant diffraction of (b) the cutoff frequency of sodium; (c) the work function. this type of light when it passes through an everyday 29. (II) Show that the energy E (in electron volts) of a photon opening, such as a door? whose wavelength is l (nm) is given by13. (I) Calculate the momentum of a photon of yellow light of 1.240 * 103 eVnm . wavelength 5.80 * 10 7 m. E = l (nm)14. (I) What is the momentum of a l = 0.014 nm X-ray photon? Use at least 4 significant figures for values of h, c, e (see15. (I) For the photoelectric effect, make a table that shows inside front cover). expected observations for a particle theory of light and for a wave theory of light. Circle the actual observed effects. *275 Compton Effect (See Section 273.) *30. (I) A high-frequency photon is scattered off of an electron16. (II) About 0.1 eV is required to break a hydrogen bond in and experiences a change of wavelength of 1.7 * 104 nm. a protein molecule. Calculate the minimum frequency and At what angle must a detector be placed to detect the scattered maximum wavelength of a photon that can accomplish this. photon (relative to the direction of the incoming photon)?17. (II) What minimum frequency of light is needed to eject electrons from a metal whose work function is 4.8 * 1019 J? Problems 799*31. (II) The quantity hmc, which has the dimensions of length, 279 Electron Microscope is called the Compton wavelength. Determine the Compton 47. (II) What voltage is needed to produce electron wavelengths wavelength for (a) an electron, (b) a proton. (c) Show of 0.26 nm? (Assume that the electrons are nonrelativistic.) that if a photon has wavelength equal to the Compton 48. (II) Electrons are accelerated by 2850 V in an electron wavelength of a particle, the photons energy is equal to the microscope. Estimate the maximum possible resolution of rest energy of the particle, mc2. the microscope.*32. (II) X-rays of wavelength l = 0.140 nm are scattered from carbon. What is the expected Compton wavelength 2711 and 2712 Spectra and the Bohr Model shift for photons detected at angles (relative to the inci- 49. (I) For the three hydrogen transitions indicated below, with dent beam) of exactly (a) 45, (b) 90, (c) 180? n being the initial state and n being the final state, is the transition an absorption or an emission? Which is higher, the276 Pair Production initial state energy or the final state energy of the atom? Finally, 33. (I) How much total kinetic energy will an electronpositron which of these transitions involves the largest energy photon? pair have if produced by a 3.64-MeV photon? (a) n = 1, n = 3; (b) n = 6, n = 2; (c) n = 4, n = 5. 34. (II) What is the longest wavelength photon that could 50. (I) How much energy is needed to ionize a hydrogen atom produce a protonantiproton pair? (Each has a mass of in the n = 3 state? 1.67 * 1027 kg. ) 51. (I) The second longest wavelength in the Paschen series in 35. (II) What is the minimum photon energy needed to hydrogen (Fig. 2729) corresponds to what transition? produce a m m pair? The mass of each m (muon) is 52. (I) Calculate the ionization energy of doubly ionized 207 times the mass of an electron. What is the wavelength lithium, Li2, which has Z = 3 (and is in the ground state). of such a photon? 53. (I) (a) Determine the wavelength of the second Balmer 36. (II) An electron and a positron, each moving at line (n = 4 to n = 2 transition) using Fig. 2729. Deter- 3.0 * 105 ms, collide head on, disappear, and produce mine likewise (b) the wavelength of the second Lyman line two photons, each with the same energy and momentum and (c) the wavelength of the third Balmer line. moving in opposite directions. Determine the energy and 54. (I) Evaluate the Rydberg constant R using the Bohr model momentum of each photon. (compare Eqs. 279 and 2716) and show that its value is 37. (II) A gamma-ray photon produces an electron and a R = 1.0974 * 107 m1. (Use values inside front cover to positron, each with a kinetic energy of 285 keV. Determine 5 or 6 significant figures.) the energy and wavelength of the photon. 55. (II) What is the longest wavelength light capable of ionizing a hydrogen atom in the ground state?278 Wave Nature of Matter 56. (II) What wavelength photon would be required to ionize 38. (I) Calculate the wavelength of a 0.21-kg ball traveling at a hydrogen atom in the ground state and give the ejected 0.10 ms. electron a kinetic energy of 11.5 eV? 39. (I) What is the wavelength of a neutron Am = 57. (II) In the Sun, an ionized helium AHe B atom makes a 1.67 * 1027 kgB traveling at 8.5 * 104 ms? transition from the n = 6 state to the n = 2 state, emit- 40. (II) Through how many volts of potential difference must ting a photon. Can that photon be absorbed by hydrogen an electron, initially at rest, be accelerated to achieve a atoms present in the Sun? If so, between what energy wavelength of 0.27 nm? states will the hydrogen atom transition occur? 41. (II) Calculate the ratio of the kinetic energy of an electron 58. (II) Construct the energy-level diagram for the He ion to that of a proton if their wavelengths are equal. Assume (like Fig. 2729). that the speeds are nonrelativistic. 59. (II) Construct the energy-level diagram for doubly ionized 42. (II) An electron has a de Broglie wavelength lithium, Li2 . l = 4.5 * 1010 m. (a) What is its momentum? (b) What 60. (II) Determine the electrostatic potential energy and the is its speed? (c) What voltage was needed to accelerate it kinetic energy of an electron in the ground state of the from rest to this speed? hydrogen atom. 43. (II) What is the wavelength of an electron of energy 61. (II) A hydrogen atom has an angular momentum of (a) 10 eV, (b) 100 eV, (c) 1.0 keV? 5.273 * 1034 kgm2s. According to the Bohr model, what 44. (II) Show that if an electron and a proton have the same is the energy (eV) associated with this state? nonrelativistic kinetic energy, the proton has the shorter 62. (II) An excited hydrogen atom could, in principle, have a wavelength. radius of 1.00 cm. What would be the value of n for a Bohr 45. (II) Calculate the de Broglie wavelength of an electron if it orbit of this size? What would its energy be? is accelerated from rest by 35,000 V as in Fig. 272. Is it rela- 63. (II) Is the use of nonrelativistic formulas justified in the tivistic? How does its wavelength compare to the size of the Bohr atom? To check, calculate the electrons velocity, v, neck of the tube, typically 5 cm? Do we have to worry in terms of c, for the ground state of hydrogen, and then about diffraction problems blurring the picture on the CRT calculate 21 - v2c2 . screen? 64. (III) Show that the magnitude of the electrostatic potential 46. (III) A Ferrari with a mass of 1400 kg approaches a freeway energy of an electron in any Bohr orbit of a hydrogen atom underpass that is 12 m across. At what speed must the car is twice the magnitude of its kinetic energy in that orbit. be moving, in order for it to have a wavelength such that 65. (III) Suppose an electron was bound to a proton, as in the it might somehow diffract after passing through this hydrogen atom, but by the gravitational force rather than single slit? How do these conditions compare to normal by the electric force. What would be the radius, and energy, freeway speeds of 30 ms? of the first Bohr orbit?

800 CHAPTER 27 Early Quantum Theory and Models of the Atom

General Problems66. The Big Bang theory (Chapter 33) states that the beginning 77. In some of Rutherfords experiments (Fig. 2719) the of the universe was accompanied by a huge burst of a particles Amass = 6.64 * 1027 kgB had a kinetic energy photons. Those photons are still present today and make up of 4.8 MeV. How close could they get to the surface of a the so-called cosmic microwave background radiation. gold nucleus (radius L 7.0 * 1015 m, charge = 79e)? The universe radiates like a blackbody with a temperature Ignore the recoil motion of the nucleus. today of about 2.7 K. Calculate the peak wavelength of this 78. By what fraction does the mass of an H atom decrease radiation. when it makes an n = 3 to n = 1 transition?67. At low temperatures, nearly all the atoms in hydrogen gas 79. Calculate the ratio of the gravitational force to the electric will be in the ground state. What minimum frequency photon force for the electron in the ground state of a hydrogen is needed if the photoelectric effect is to be observed? atom. Can the gravitational force be reasonably ignored?68. A beam of 72-eV electrons is scattered from a crystal, as 80. Electrons accelerated from rest by a potential difference in X-ray diffraction, and a first-order peak is observed at of 12.3 V pass through a gas of hydrogen atoms at room u = 38. What is the spacing between planes in the temperature. What wavelengths of light will be emitted? diffracting crystal? (See Section 2511.) 81. In a particular photoelectric experiment, a stopping potential of 2.10 V is measured when ultraviolet light of69. A microwave oven produces electromagnetic radiation at wavelength 270 nm is incident on the metal. Using the l = 12.2 cm and produces a power of 720 W. Calculate the same setup, what will the new stopping potential be if blue number of microwave photons produced by the microwave light of wavelength 440 nm is used, instead? oven each second. 82. Neutrons can be used in diffraction experiments to probe70. Sunlight reaching the Earths atmosphere has an intensity of the lattice structure of crystalline solids. Since the neutrons about 1300 Wm2 . Estimate how many photons per square wavelength needs to be on the order of the spacing meter per second this represents. Take the average wave- between atoms in the lattice, about 0.3 nm, what should length to be 550 nm. the speed of the neutrons be?71. A beam of red laser light (l = 633 nm) hits a black wall 83. In Chapter 22, the intensity of light striking a surface was and is fully absorbed. If this light exerts a total force related to the electric field of the associated electro- F = 5.8 nN on the wall, how many photons per second are magnetic wave. For photons, the intensity is the number of hitting the wall? photons striking a 1-m2 area per second. Suppose 1.0 * 1012 photons of 497-nm light are incident on a 1-m272. A flashlight emits 2.5 W of light. As the light leaves the surface every second. What is the intensity of the light? flashlight in one direction, a reaction force is exerted on Using the wave model of light, what is the maximum the flashlight in the opposite direction. Estimate the size of electric field of the electromagnetic wave? this reaction force. 84. The intensity of the Suns light in the vicinity of the Earth73. A photomultiplier tube (a very sensitive light sensor), is is about 1350 Wm2. Imagine a spacecraft with a mirrored based on the photoelectric effect: incident photons strike a square sail of dimension 1.0 km. Estimate how much thrust metal surface and the resulting ejected electrons are (in newtons) this craft will experience due to collisions with collected. By counting the number of collected electrons, the Suns photons. [Hint: Assume the photons bounce off the the number of incident photons (i.e., the incident light sail with no change in the magnitude of their momentum.] intensity) can be determined. (a) If a photomultiplier tube 85. Light of wavelength 280 nm strikes a metal whose work is to respond properly for incident wavelengths through- function is 2.2 eV. What is the shortest de Broglie wave- out the visible range (410 nm to 750 nm), what is the length for the electrons that are produced as photoelectrons? maximum value for the work function W0 (eV) of its metal surface? (b) If W0 for its metal surface is above a certain 86. Photons of energy 6.0 eV are incident on a metal. It is threshold value, the photomultiplier will only function for found that current flows from the metal until a stopping incident ultraviolet wavelengths and be unresponsive to potential of 3.8 V is applied. If the wavelength of the visible light. Determine this threshold value (eV). incident photons is doubled, what is the maximum kinetic energy of the ejected electrons? What would happen if the74. If a 100-W lightbulb emits 3.0% of the input energy as wavelength of the incident photons was tripled? visible light (average wavelength 550 nm) uniformly in all 87. What would be the theoretical limit of resolution for an directions, estimate how many photons per second of electron microscope whose electrons are accelerated visible light will strike the pupil (4.0 mm diameter) of the through 110 kV? (Relativistic formulas should be used.) eye of an observer, (a) 1.0 m away, (b) 1.0 km away. 88. Assume hydrogen atoms in a gas are initially in their75. An electron and a positron collide head on, annihilate, and ground state. If free electrons with kinetic energy 12.75 eV create two 0.85-MeV photons traveling in opposite direc- collide with these atoms, what photon wavelengths will be tions. What were the initial kinetic energies of electron emitted by the gas? and positron? 89. Visible light incident on a diffraction grating with slit76. By what potential difference must (a) a proton Am = spacing of 0.010 mm has the first maximum at an angle of 1.67 * 1027 kgB, and (b) an electron Am = 9.11 * 1031 kgB, 3.6 from the central peak. If electrons could be diffracted be accelerated from rest to have a wavelength by the same grating, what electron velocity would produce l = 4.0 * 1012 m? the same diffraction pattern as the visible light?

General Problems 801

90. (a) Suppose an unknown element has an absorption spec- 93. An electron accelerated from rest by a 96-V potential trum with lines corresponding to 2.5, 4.7, and 5.1 eV above difference is injected into a 3.67 * 104 T magnetic field its ground state and an ionization energy of 11.5 eV. Draw where it travels in an 18-cm-diameter circle. Calculate em an energy level diagram for this element. (b) If a 5.1-eV from this information. photon is absorbed by an atom of this substance, in which 94. Estimate the number of photons emitted by the Sun in a state was the atom before absorbing the photon? What will year. (Take the average wavelength to be 550 nm and the be the energies of the photons that can subsequently be intensity of sunlight reaching the Earth (outer atmosphere) emitted by this atom? as 1350 Wm2 .)91. A photon of momentum 3.53 * 1028 kg ms is emitted 95. Apply Bohrs assumptions to the EarthMoon system to from a hydrogen atom. To what spectrum series does this calculate the allowed energies and radii of motion. Given photon belong, and from what energy level was it ejected? the known distance between the Earth and Moon, is the92. Light of wavelength 464 nm falls on a metal which has a quantization of the energy and radius apparent? work function of 2.28 eV. (a) How much voltage should be 96. At what temperature would the average kinetic energy applied to bring the current to zero? (b) What is the (Chapter 13) of a molecule of hydrogen gas (H 2) be suffi- maximum speed of the emitted electrons? (c) What is the cient to excite a hydrogen atom out of the ground state? de Broglie wavelength of these electrons?

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1. Name the person or people who did each of the following: 5. (a) From Sections 223, 244, and 273, estimate the mini- (a) made the first direct measurement of the charge-to- mum energy (eV) that initiates the chemical process on the mass ratio of the electron (Section 271); (b) measured retina responsible for vision. (b) Estimate the threshold the charge on the electron and showed that it is quantized photon energy above which the eye registers no sensation (Section 271); (c) proposed the radical assumption that of sight. the vibrational energy of molecules in a radiating object is 6. (a) A rubidium atom (m = 85 u) is at rest with one electron quantized (Sections 272, 273); (d) found that light in an excited energy level. When the electron jumps to (X-rays) scattered off electrons in a material will decrease the ground state, the atom emits a photon of wavelength the energy of the photons (Section 275); (e) proposed l = 780 nm. Determine the resulting (nonrelativistic) that the wavelength of a material particle would be related recoil speed v of the atom. (b) The recoil speed sets the to its momentum in the same way as for a photon lower limit on the temperature to which an ideal gas of (Section 278); (f ) performed the first crucial experiment rubidium atoms can be cooled in a laser-based atom trap. illustrating electron diffraction (Section 278); (g) deciphered Using the kinetic theory of gases (Chapter 13), estimate the nuclear model of the atom by aiming a particles at this lowest achievable temperature. gold foil (Section 2710). 7. Suppose a particle of mass m is confined to a one- 2. State the principle of complementarity, and give at least two dimensional box of width L. According to quantum theory, experimental results that support this principle for electrons the particles wave (with l = hmv ) is a standing wave and for photons. (See Section 277 and also Sections 273 with nodes at the edges of the box. (a) Show the possible and 278.) modes of vibration on a diagram. (b) Show that the kinetic 3. Imagine the following Youngs double-slit experiment energy of the particle has quantized energies given by using matter rather than light: electrons are accelerated ke = n2h28mL2, where n is an integer. (c) Calculate the through a potential difference of 12 V, pass through two ground-state energy (n = 1) for an electron confined to a closely spaced slits separated by a distance d, and create box of width 0.50 * 1010 m. (d) What is the ground-state an interference pattern. (a) Using Example 2711 and energy, and speed, of a baseball (m = 140 g) in a box Section 243 as guides, find the required value for d if the 0.65 m wide? (e) An electron confined to a box has a first-order interference fringe is to be produced at an angle ground-state energy of 22 eV. What is the width of the box? of 10. (b) Given the approximate size of atoms, would it [Hint: See Sections 278, 2713, and 1112.] be possible to construct the required two-slit set-up for this experiment? 4. Does each of the following support the wave nature or the particle nature of light? (a) The existence of the cutoff frequency in the photoelectric effect; (b) Youngs double- slit experiment; (c) the shift in the photon frequency in Compton scattering; (d) the diffraction of light.